Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 + 8z + 7}{z^2 - 9z} \times \dfrac{z - 9}{z + 1} $
Solution: First factor the quadratic. $q = \dfrac{(z + 1)(z + 7)}{z^2 - 9z} \times \dfrac{z - 9}{z + 1} $ Then factor out any other terms. $q = \dfrac{(z + 1)(z + 7)}{z(z - 9)} \times \dfrac{z - 9}{z + 1} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (z + 1)(z + 7) \times (z - 9) } { z(z - 9) \times (z + 1) } $ $q = \dfrac{ (z + 1)(z + 7)(z - 9)}{ z(z - 9)(z + 1)} $ Notice that $(z - 9)$ and $(z + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(z + 1)}(z + 7)(z - 9)}{ z(z - 9)\cancel{(z + 1)}} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $q = \dfrac{ \cancel{(z + 1)}(z + 7)\cancel{(z - 9)}}{ z\cancel{(z - 9)}\cancel{(z + 1)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac{z + 7}{z} ; \space z \neq -1 ; \space z \neq 9 $